A tennis ball is thrown from a cliff at an angle 31 ∘
 above the horizontal. The tennis ball hits the ground 3.7 s
 later at a horizontal distance of 153 ft
 from where it was thrown. What is the maximum height above the ground that the ball reaches?

We can approach this problem using the kinematic equations for projectile motion.

### Given:
- Time of flight: \( t = 3.7 \) s
- Horizontal distance (range): \( x = 153 \) ft
- Angle of projection: \( \theta = 31^\circ \)
- Gravity: \( g = 32.2 \, \text{ft/s}^2 \) (acceleration due to gravity)

We aim to find the **maximum height** of the ball above the ground.

### Step 1: Horizontal Motion
The horizontal velocity \( v_x \) is constant, and the horizontal distance traveled is given by:

\[
x = v_x \cdot t
\]

The horizontal velocity \( v_x \) is related to the initial velocity \( v_0 \) and the angle of projection \( \theta \) as:

\[
v_x = v_0 \cdot \cos(\theta)
\]

Substituting into the equation for the range:

\[
153 = v_0 \cdot \cos(31^\circ) \cdot 3.7
\]

We can solve for \( v_0 \).

### Step 2: Vertical Motion
The vertical component of the initial velocity is:

\[
v_y = v_0 \cdot \sin(\theta)
\]

The maximum height is reached when the vertical velocity becomes zero. Using the kinematic equation:

\[
v_y^2 = v_{0y}^2 - 2g h_{\text{max}}
\]

At the maximum height, \( v_y = 0 \), so:

\[
0 = v_0^2 \cdot \sin^2(\theta) - 2g h_{\text{max}}
\]

Solving for \( h_{\text{max}} \):

\[
h_{\text{max}} = \frac{v_0^2 \cdot \sin^2(31^\circ)}{2g}
\]

We will first solve for \( v_0 \) from Step 1, then calculate \( h_{\text{max}} \). Let's compute.The initial velocity of the tennis ball is approximately \( 48.24 \, \text{ft/s} \).

The maximum height the ball reaches above the point it was thrown from is approximately \( 9.59 \, \text{ft} \).

If you'd like further details or have any questions, feel free to ask!

Here are some related questions to explore:
1. How do the horizontal and vertical components of the velocity affect the motion?
2. What would happen if the angle of projection was increased?
3. How does air resistance affect projectile motion, and why is it often ignored in such calculations?
4. Can you calculate the time taken to reach the maximum height?
5. What is the velocity of the tennis ball when it hits the ground?

**Tip:** When dealing with projectile motion, it's useful to break down the motion into horizontal and vertical components to simplify calculations.

A tennis ball is thrown from a cliff at an angle 31° above the horizontal. The tennis ball hits the ground 3.7 s later at a horizontal distance of 153 ft from where it was thrown. What is the maximum height above the ground that the ball reaches?

Solve the problem of finding the maximum height of a tennis ball thrown from a cliff at an angle of 31° above the horizontal. The ball hits the ground after 3.7 seconds, covering a horizontal distance of 153 ft. This solution uses kinematic equations to determine the initial velocity and maximum height, suitable for Grades 11-12.

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